As I understand it, a bridge rectifier is a set of diodes that play a trick of logic that keeps the polarity positive.

So, within the context of a household DC transformer does it matter what the frequency of the AC wave is?

I guess it does to a point, since the DC output has a ripple and the smoothing needs to make-up the dips, so I imagine there is some kind of acceptable range, probably dictated by the reservoir capacitor.

1 Answer
1

There are three issues at work here: the voltage ripple on the output, peak currents through the rectifier, and the reverse-recovery losses of the diodes. Within the context you specified (household) the latter two are unlikely to matter, but I'll include them anyway for completeness.

First, the ripple voltage. I=C dV/dt. If you know the current load (I), and you know how much ripple is acceptable for your application (dV), then you can extract a relationship between the capacitance (C) and the half-period of your AC line (dt). How much ripple is acceptable? Depends on the application. But for a fixed capacitor, higher frequencies will reduce the size of the ripple, and lower frequencies will increase it.

Second, peak currents. The rectifier doesn't conduct all the time; it only conducts when the AC wave is higher than the value of the DC capacitor. So your AC voltage looks like a sine wave, but the current being pulled looks like a big spike just at the peak of the wave.

Now, these spikes are sub-optimal. They don't look at all like a sine wave, so they cause harmonics on the AC line. And the RMS current of those spikes is much higher than a sine wave of equivalent power delivery would have, so they stress any fuses or breakers upstream.

Characterizing the current spike can get complicated, because it depends on the frequency of the AC, the capacitor, and the inductance of the AC line. The more the inductance, the wider-in-time and shorter-in-amplitude the pulses get. (For high-power three-phase applications it's common to add a big inductor to purposefully spread the diode conduction time and reduce all those problems, but I don't think it's common on household stuff. In general, people don't really care about harmonics in those contexts.) But unless you're pulling something close to the full rated power of the breaker, this won't be much of an issue.

Third, the diodes involved usually have a reverse-recovery time. When a diode goes from being forward-biased to being reverse-biased, it actually takes a finite time for it to stop conducting. (There are zero-recovery diodes, but they aren't usually used for 60 Hz work.) During that time, the diode acts like a short circuit, meaning it dissipates a lot of power. This time is typically on the order of microseconds, so for a 60 Hz line you don't see much additional loss, and can probably ignore recovery losses. If you were operating in kilohertz, you'd have to account for it.

Actually between the two half periods there is a small gap (Δt) because of the forward voltage drop of the diodes. In low output voltage this gap plays important role in current capability of the bridge.
–
GR TechMar 31 '14 at 13:29

1

What actually matters more than the bridge rectifier, apart from sag based on the decoupling caps, (even though the asker's question doesn't ask about this) is the effect of frequency on the transformer. Transformers are less efficient at lower frequencies, so the transformer will heat up more when used at a lower than specced frequency, and could even saturate if particularly underspecced.
–
nitro2k01Mar 31 '14 at 18:10

The RMS current of those spikes is much higher than a sine wave of equivalent power delivery would have, so they stress any fuses or breakers upstream. -- I dispute that assertion. While it is true that the increase in current (and subsequent rollof) are sudden, the actual current draw will never be more than the highest point of the original sine wave.
–
Robert HarveyMar 31 '14 at 18:21

1

@RobertHarvey I'm not sure what you mean by "the highest point of the original sine wave". There's no sinusoidal current to compare against in this system, I don't think. What I'm saying is that if you take the average power delivery of that narrow spike, and then you compare it against a sinusoidal current of equivalent power delivery, the RMS of the spikes will be higher.
–
Stephen CollingsMar 31 '14 at 20:25

@StephenCollings: A sinusoidal current of equivalent power delivery is going to be of much lower amplitude than the sine wave pictured above, so I don't see how it's going to stress breakers unless they are somehow sensitive to the induced harmonics. If anything, it's going to reduce breaker stress. Power supplies are not the only thing that exhibit this characteristic (of changing the duty cycle of the AC load); so do light dimming circuits.
–
Robert HarveyMar 31 '14 at 20:28